I believe Cardano's method used to be taught in high school, but before my birth. Consequently, the clearest explanation I have found is in the "elementary algebra" book by Euler, from which I adapted this presentation for bright young children, like your granddaughter, in a summer camp. All it requires is a good understanding of solving quadratics. This is from notes on my website, a little ways into them, below problem #13.
https://www.math.uga.edu/sites/default/files/inline-files/epsilon13.pdf
Next I want give Euler's explanation of how to solve cubic equations.
first he shows that any cubic equation can be transformed by a trick to change the cubic into one with no X^2 term. So it is only necessary to be able to solve cubics like this one:
X^3=pX+q. E.g.suppose that we have X^3=9X+28.
Then Euler explains that to solve this all we need to do is find two numbers u,v such that 3uv=9 and u^3+v^3=28. Then X=u+v will solve the cubic.
14. See if you can use Euler's method to solve X^3 = 9X + 28.
(Essentially this solution method was found apparently by Scipio del Ferro, and later Tartaglia, who explained it to Girolamo Cardano, who eventually published it. It is often called "Cardano's method".)
15. Try this one as well: X^3 = -18X + 19.
Solving cubics used to be taught in elementary algebra books, but when I went to high school, it was no longer done, and I think is not commonly done now. This is another reason to prefer the great old algebra books like those of Euler and LaGrange.
16. Show this method can be used on all cubics of form X^3 = pX + q.
I.e. i) show that one can always find numbers u,v such that u^3+v^3 = q, and 3uv = p.
(Hint: let A=u^3 and B=v^3. Then you know that A+B = q, and AB = p^3/27; why? Then show how to find A and B, and then tell how to find u and v.)
ii) Show that if X=u+v and if 3uv=p and u^3+v^3=q,thenX^3=pX +q.
The whole point of understanding quadratic equations is this. Write them like X^2-bX+c = 0, and then b is always the sum of the solutions and c is always their product.
Vice versa, whenever you are looking for two numbers and you already know their sum and their product, then you can always find the numbers as the solutions of a quadratic equation.
The whole point of solving cubics is this:
First we know from studying quadratics that we can always find two numbers whose sum and product are known.
Next Euler shows that to solve the cubic X^3 = pX + q, all you need is two numbers u,v such that u^3 +v^3 = q, and 3uv = p.
So we just need to find u and v. But it turns out we already know how to find their cubes, u^3 and v^3.
I.e. what do we know about the cubes of u,v? we know their sum u^3 + v^3 = q, and we know their product, since 3uv = p, so 27 u^3v^3 =p^3, hence u^3v^3 = p^3/27.
So since we know the sum and product of u^3 and v^3, we can find u^3 and v^3 by solving a quadratic! Then we can take cube roots to find u and v.
I.e. if r is a root of the quadratic t^2 - qt + (p^3/27), set u = cuberoot(r) and v = p/3u, and then u + v = X solves the cubic!Unfortunately, it seems the given problem is not well suited to this method, but she might enjoy using it on #14 and #15 given here. Of course, the first step of Euler's method, substituting Y = X-1, or X = Y-1, and expanding yields the simpler equation Y^3 - Y -6 = 0, whose solutions are each one less than the solutions of the original equation, and 2 is an obvious solution here, but the other solutions are solutions of the quadratic Y^2 + 2Y + 3. A moral is perhaps that Cardano's formula is not always that useful, in comparison to the rational root theorem.
